$$\text{Take the sequence, } x_n = \frac {2n^2} {n+1}$$
It clearly doesn't converge since the numerator will always be bigger than the denominator, but how does one formally prove this?
We know that if a sequence converges, then $|x_n - x| < \epsilon$ . If we let $\epsilon = 1$, then we can construct,
$$|\frac {2n^2} {n+1} -x| <1$$ But how would one prove that this inequality doesn't hold?
Thanks.
It suffices to show that it's not bounded. $\forall M \in \mathbb{R}$, let $N \in \mathbb{Z}^+$ be the smallest positive integer such that $N > M$. Then, $\forall n \geq N$: $$ x_n = \frac{2n^2}{n + 1} \geq \frac{2n^2}{n + n} = n \geq N > M $$ Since every convergent sequence is bounded, $(x_n)$ does not converge.